Momentum problem with collision

First of all, you missed a sign.

 

Vy is positive, and that is why you have to put the negative sign in front.

Although you haven't explicitly written it down, it's the components of the velocity vectors that you are equating.

So, you're second equation is something like
0j=2.1sin(theta)j-Vyj

 

By j, I meant a unit-vector in the direction parallel to the y-axis.

0j=2.1sin(theta)j-Vyj
=> 0j=(2.1sin(theta)-Vy)j

Since the vectors are equal, so are their magnitudes, and that's how you get
0=2.1sin(theta)-Vy

 

hey guys may b u shud consider forming equs fom the line of collision.....tats how they did in our class....im not v comfertable wit it either, but ill get ma note book if u want

 

well, in a similar prob, v got the first equs by e=v(seperation)/v(approach)...the other relations were used on the basis of LOC

 

yea ...sumthing like tat...shud i get ma notebook?

 

e=co-eff of restitution, n LOC is line of collision, , it says here newtons law of collision can only b applied along the loc

 

ummm...u shud know tat im living in bombay, so even if the laws of physics dont change, the methods of attacking a ques do change ....n im currently trainig for iit entrance exam...so, idk wht level u r at

 

now, for any spherrical bodies, loc is the line joining the centres of the two coliding bodies......u take the components of the resulting velocities along the loc, get the equs n solve......now i can giv u the solution inna while, but my personal advise wud b ,dont listen to me atall if the concept of loc is entirely new to u.....but frankly, i dont c how the prob can b tackled otherwise

 

not quite, u on msn?

 

dear, u just said tat its a elastic collision, n to avoid friction n all, u ment a perfectly elastic collision, in this case, e=1, always

 

whcih brings us to the eqn u already hav, u1= v1cos(theta)- v2....the udder eqs, i recommend u get aquainted wit the concept of loc